A man can row at 5 km/hr in still water, if the river is running at 1 km/hr it takes him 75 minutes to row to a place and back. How far is the place?
Answer: B
Speed downstream = (5+1)km/hr = 6 km/hr
Speed upstream = (5-1)km/hr = 4 km/hr
Let the required distance be x km.
Then x/6 + x/4 = 75/60 = 5/4 => (2x + 3x) =15 => x =3.
Required distance = 3 km
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A boat goes 6 km in an hour in still water. It takes thrice as much time in covering the same distance against the current. Speed of the current is?
Answer: C
With \( b \) the still-water speed and \( r \) the stream's,
\[ \text{downstream} = b + r, \qquad \text{upstream} = b - r \]
Still water gives \( b = 6 \) km/hr. Taking three times as long against the current means going at one third the speed:
\[ b - r = \frac{6}{3} = 2 \text{ km/hr} \]
\[ r = 6 - 2 = 4 \text{ km/hr} \]
The comparison here is with the still-water time, not the downstream time - so \( b - r \) is compared with \( b \), giving \( r = \tfrac{2b}{3} \).
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A boat covers a certain distance downstream in 1 hour and returns in 1 hour 30 minutes. If the stream flows at 3 km/hr, what is the speed of the boat in still water ?
Answer: C
Correct answer: 15 km/hr
The distance is the same each way, so speed multiplied by time must match:
\[(b + 3) \times 1 = (b - 3) \times 1.5\]
\[b + 3 = 1.5b - 4.5 \quad\Longrightarrow\quad 7.5 = 0.5b \quad\Longrightarrow\quad b = 15 \text{ km/hr}\]
The distance is then \(18 \times 1 = 18\) km, and the return at 12 km/hr indeed takes 1.5 hours.
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A boat is rowed downstream at 15.5km/hr and upstream at 8.5 km/hr. The speed of the stream is?
Answer: A
With \( b \) the still-water speed and \( r \) the stream's,
\[ \text{downstream} = b + r, \qquad \text{upstream} = b - r \]
\[ b = \frac{(b+r) + (b-r)}{2}, \qquad r = \frac{(b+r) - (b-r)}{2} \]
\[ r = \frac{15.5 - 8.5}{2} = \frac{7}{2} = 3.5 \text{ km/hr} \]
The boat does 12 km/hr in still water. The 7 km/hr option is the raw difference, which is twice the current.
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A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is:
Answer: B
Write \( b \) for the boat's speed in still water and \( r \) for the stream's. Then
\[ \text{downstream} = b + r, \qquad \text{upstream} = b - r \]
\[ 4 \text{ h } 30 \text{ min} = 4.5 \text{ h} \]
\[ \frac{30}{15 + r} + \frac{30}{15 - r} = 4.5 \]
Combining the fractions over the common denominator \( (15+r)(15-r) = 225 - r^{2} \):
\[ \frac{30(15 - r) + 30(15 + r)}{225 - r^{2}} = \frac{900}{225 - r^{2}} = 4.5 \]
\[ 225 - r^{2} = 200 \quad\Longrightarrow\quad r^{2} = 25 \quad\Longrightarrow\quad r = 5 \text{ km/h} \]
The \( r \) terms in the numerator cancel, which is what turns a messy equation into a simple one.
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A man's speed with the current is 20 kmph and speed of the current is 3 kmph. The Man's speed against the current will be
Answer: C
Speed with current is 20,
speed of the man + It is speed of the current
Speed in still water = 20 - 3 = 17
Now speed against the current will be
speed of the man - speed of the current
= 17 - 3 = 14 kmph
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A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is:
Answer: B
Write \( b \) for the boat's speed in still water and \( r \) for the stream's. Then
\[ \text{downstream} = b + r, \qquad \text{upstream} = b - r \]
\[ 4 \text{ h } 30 \text{ min} = 4.5 \text{ h} \]
\[ \frac{30}{15 + r} + \frac{30}{15 - r} = 4.5 \]
Combining the fractions over the common denominator \( (15+r)(15-r) = 225 - r^{2} \):
\[ \frac{30(15 - r) + 30(15 + r)}{225 - r^{2}} = \frac{900}{225 - r^{2}} = 4.5 \]
\[ 225 - r^{2} = 200 \quad\Longrightarrow\quad r^{2} = 25 \quad\Longrightarrow\quad r = 5 \text{ km/h} \]
The \( r \) terms in the numerator cancel, which is what turns a messy equation into a simple one.
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A man rows 750 m in 675 seconds against the stream and returns in 7 and half minutes. His rowing speed in still water is
Answer: B
Rate upstream = (750/675) = 10/9 m/sec
Rate downstream (750/450) m/sec = 5/3 m/sec
Rate in still water = (1/2)*[(10/9) + (5/3)] m/sec.
= 25/18 m/sec
= (25/18) × (18/5) kmph
= 5 kmph
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A boat takes 90 minutes less to travel 36 km downstream than to travel the same distance upstream. If the speed of the stream is 2 km/hr, what is the speed of the boat in still water ?
Answer: C
Correct answer: 10 km/hr
With \(b\) the still-water speed, the two times differ by \(1\tfrac{1}{2}\) hours:
\[\frac{36}{b-2} - \frac{36}{b+2} = 1.5\]
\[36 \cdot \frac{(b+2)-(b-2)}{b^2 - 4} = 1.5 \quad\Longrightarrow\quad \frac{144}{b^2 - 4} = 1.5\]
\[b^2 - 4 = 96 \quad\Longrightarrow\quad b^2 = 100 \quad\Longrightarrow\quad b = 10 \text{ km/hr}\]
Checking: 36 km takes 3 hours downstream at 12 km/hr and 4.5 hours upstream at 8 km/hr.
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A boat running upstream takes 8 hours 48 min to cover a certain distance while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and the speed of water current respectively?
Answer: C
Let the speed of boat be x km/hr and speed of stream be ykm/hr.
44/5 × (x -y) = 4x (x +y)
=> 44(x-y) = 20 (x+y)
=>11 (x-y) = 5 (x +y)
=> 6x =16 y
=> x/y = 16/6 = 8/3
=> x : y = 8 : 3
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